Unclued, I think you must have gone wrong somewhere, because the sum of the 20 digits in the grid is 98.
With having to encode the four 2-digit answers, this puzzle involved me in a lot of brute computation where I tried all possible combinations in some places. You are correct that looking at 9ac and 6dn is a good starting point. As the code number has only two digits, the third digit in the top row of the code-square must be 1, 2 or 3. However, if you encode the code number, which is the first and second digits in the code-square, then this is encoded as the second and third digits. Hence the bottom right hand cell in the grid must be 1, 2 or 3. But as 6dn is a square, it can’t end in 2 or 3, so it must end in 1.
There are only five three-digit squares ending in 1, and all of them have an even number as their middle digit. Because the middle digit of 6dn is the last digit of 7ac, this means that 7ac is even, and as 3dn is a multiple of 7ac, 3dn is also even. Looking at the intersection of 2ac and 3dn and noting that 3dn is the reverse of 2ac leaves only a restricted choice of numbers for 2ac (particularly as 2ac is a multiple of another grid entry and so cannot be prime).
There is a very restricted number of combinations for 2ac, 3dn, 7ac and 6dn that satisfy all the requirements, and there are therefore very few possible answers for 2dn before this is encoded. I tried all possible Playfair code-squares (there are 56 of them, as the first digit can in theory be any of the eight digits from 2 to 9 (zero is specifically excluded and 1 can’t be in the code number because it’s already accounted for as the third digit in the top row of the code-square), and the second digit has to be a different digit, so there are 8 x 7 = 56 possibilities. Although I was able quite quickly to reduce the possible code numbers to only two of these 56 possibilities, I didn’t confirm the actual code number until right at the end, solving 1ac and 1dn together. Another route might have got there more quickly, however.